Example: Cosine rule to find a length

First we match up the information in the question to the letters in the formula.

a=x, A = 44\degree, b=5 and c=7.

So, we’re ready to substitute the values into the formula.

x^2=5^2+7^2-(2\times5\times7\times\cos(44))=25+49-70\cos(44)

Taking the square root of both sides, and putting it into the calculator, we get

x=\sqrt{25+49-70\cos(44)}=4.9\text{ (1dp)}.

Example Questions

1) Use the cosine rule to find the side-length marked x below to 2sf.

Firstly, we need appropriately label the sides of this triangle. Firstly, we set a=x, and therefore we get that A=19, since it is the angle opposite. It doesn’t matter how we label the other two sides, so here we’ll let b=86 and c=65.

Then, taking the square root, and putting it into the calculator, we get

x=\sqrt{7,396+4,225-11,180\cos(19)}=32\text{cm (2sf)}

2) Use the cosine rule to find the angle marked x below to 1dp.

As always, we must label our triangle. Firstly, assign the thing we’re looking for to be a=x, and therefore make the side opposite to it is A=6. Then, it doesn’t matter how we choose the other two sides, so we will let b=5 and c=7.

Here, we will use the rearranged version of the formula that looks like

\cos A=\dfrac{b^2+c^2-a^2}{2bc},

So, subbing these values into the equation, we get

\cos x=\dfrac{5^2+7^2-6^2}{2\times5\times7}=\dfrac{25+49-36}{70}

Taking \cos^{-1} of both sides, and putting it into a calculator, we get